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Impulse & Impulse-Momentum Theorem

Patrick Ford
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Hey guys, so in this video, I'm going to introduce you to another physical quantity that's related to momentum that you're absolutely gonna need to know. And it's called impulse. So let's take a look here basically when an object experiences an impulse, which will use the letter J. For it experiences a change in momentum. So what we rewrite this is jay is equal to delta P. I'm gonna I'm gonna come back to this in just a second here. What I want to do is actually want to sort of rearrange this equation to be a little bit more useful. So remember that P is just M. V. And delta just means final minus initial. So really delta P is really just equal to M. V. Final minus M. V. Initial here. Now, how do we actually get from J? Two delta P. This actually is gonna be done in your textbooks. I'm going to show you this really quickly. It really actually just comes from Newton's second law, which remembers F equals M. A. But we're going to rewrite this in terms of momentum. I'm gonna show you real quickly. So we're gonna use the definition for a which remember the acceleration is just delta V over delta T. But now what I can do is these problems, the mass is always going to remain constant. So I can actually bring it inside of the delta side and that's perfectly fine. So I have delta Mv divided by delta T. Now you should know by now that actually this M. V. Is really just your momentum. So we can rewrite this and we can say that F is equal to delta P over delta T. So there's two ways we can actually now sort of express or write Newton's second law, we can say F equals M. A. But the way that Newton originally wrote it was that F is actually equal to delta P over delta T. So it's changing momentum over changing time. So what I can do here say, I actually I can actually rewrite this new expression now and I can get back to impulse. So I have F. Is not equal to Emma is just equal to delta P over delta T. And now I'm just going to move the change in time over to the other side. And when I come up with is I come up with F times delta T. Is equal to delta P. This thing here on the left side is actually what is the definition of an impulse? So J. Is really just F times delta T. The way I like to think about this is if you look at the definition for impulse, an impulse is like a very sudden thing that happens to a very short amount of time. and that's exactly what's going to happen in your problems you're gonna have in physics forces that act over some change in time and that's what an impulse is. So really what happens is we're just going to write this equation, jake was F delta T. And then you could write either one of these sort of, but I like to write like this using basically these three terms J. F delta T. And the change in the momentum. So the units that we're gonna use are either gonna be newton seconds. And it really just becomes uh that comes from force times time. Or if we're talking about changing momentum, then the units are just gonna be the same as momentum kilogram meters per second. Let's go ahead and work out a problem here. So here we have a crate that is initially at rest, so the initial velocity is zero and we're gonna push it now, we're gonna push it with this 100 m 100 newton force. That's applied force for a certain amount of time. It's gonna be eight seconds here. This is gonna be a delta T. And in part, I want to calculate the impulse that I'm delivering to the crates. So what happens here is in part, I want to calculate the impulse which is J. Remember I'm always going to write it as F delta T. And this is also related to M. V. Final minus M. V. Initial jay is really just F delta T. But it's also the change in the momentum. We'll talk about that in just a second here. So if you look through my variables or to have is I have F and delta T. So I can actually just use this first part here. This first equals sign. To calculate the impulse. So you're J is just equal to the force, which is 100 times the time which is eight. And this is equal to 800 newton seconds. And that's the answer. That's the impulse 800. So let's take a look at part B now, in part B. We want to calculate the creates speed after eight seconds and we want to use impulse to do this. So basically what happens is that once you've exerted this force over delta T. Of eight, the box is gonna be over here. And because you've pushed it over some time, it's gonna have some final velocity here. V final. How do we figure that out? Well, we're gonna write our impulse equation, jay equals F delta T. And the Sequels mv final minus M. V initial. We want to find this V final here. so we just have to figure out everything else. Well, remember that the box initially starts with an initial velocity of zero. So actually there is no initial momentum. So in the final and initial zero. Now remember that we just calculated with delta T. Is this is just 800. So we can actually say, is that 800? Which is the impulse is equal to M Times the final. So now I'm just gonna move the em over to the other side and I have that the final is equal to 800 divided by 50. And if you work this out, you're gonna get 16 m/s. All right, so that's what you sort of both sides of the impulse equation to figure this out. And all these problems, you're gonna be given three or four out of three out of these four variables F delta T. M. And some some combination of the velocities. And as long as you have three out of those four, you can always figure out the other one by using this equation here. All right, So I have one last point to make here, which is that we're actually going to see a lot of similarities between impulse and momentum and work and kinetic energy, which we've seen before. So impulse kind of relates to momentum the same way that work relates to kinetic energy. Remember that jay is equal to it's defined as the force times delta T. In the same way that the work is defined as F times delta X. And we actually used F. D. Co signed data. But the sort of simplified version is that if you have forced displacement in the same direction, F times delta X. So both of these things sort of sort of defined as forced time to change in either displacement or time. And what they cause is the F. Times delta T. Causes a change in your momentum. Right? So we saw in this problem here that this impulse causes a change in your momentum in the same way that the force actually caused a change in the kinetic energy. So there's a lot of similarities there. Um some pretty interesting things. So that's it for this one. Guys, let me know if you have any questions.